Pre-Algebra Examples

$15 = x (x - 2)$

Step 1

Rewrite the equation as $x (x - 2) = 15$ .

$x (x - 2) = 15$

Step 2

Simplify $x (x - 2)$ .

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Step 2.1

Apply the distributive property.

$x \cdot x + x \cdot - 2 = 15$

Step 2.2

Simplify the expression.

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Step 2.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 2 = 15$

Step 2.2.2

Move $- 2$ to the left of $x$ .

$x^{2} - 2 x = 15$

Step 3

Subtract $15$ from both sides of the equation.

$x^{2} - 2 x - 15 = 0$

Step 4

Factor $x^{2} - 2 x - 15$ using the AC method.

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Step 4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Step 4.2

Write the factored form using these integers.

$(x - 5) (x + 3) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x + 3 = 0$

Step 6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 8

The final solution is all the values that make $(x - 5) (x + 3) = 0$ true.

$x = 5, - 3$